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Why does thrust differ in different reference frames?
If a jet engine flies with 4 m/s in the air and we want to calculate thrust for two different reference frames, can we prove that the calculated thrust for both reference frames are equal to each other, so either of them is correct?
According to an observer on the ground;
Air velocity at inlet: 0 m/s
Air velocity at nozzle: 5 m/s
Aircraft velocity: 4 m/s (just for an example)
Air density: 1 kg/m^3
Inlet and nozzle area: 1 m^2
Inlet mass flow rate: 0 kg/s
Nozzle mass flow rate: 5 kg/s
If we convert these values to engine-based reference frame;
Air velocity at inlet: 4 m/s
Air velocity at nozzle: 9 m/s
Aircraft velocity: 0 m/s (just for an example)
Air density: 1 kg/m^3
Inlet and nozzle area: 1 m^2
Inlet mass flow rate: 4 kg/s
Nozzle mass flow rate: 9 kg/s
Can you make calculations for engine-based reference frame and ground-based reference frame and prove they are equal to each other?
According to my calculations, the ground-based reference thrust is 25 N, and the engine-based reference thrust is 81-16: 65 N.
Why do I find different thrust values for different frames? Which one is correct? In general, we make these calculations in aerospace engineering for the engine-based reference frame in which the aircraft is considered to be stationary, but why?
Thanks.

I can't give a full answer on why this is incorrect, but I believe it has something to do with not accounting for the direction of the velocity. If the aircraft is moving at 4m/s (say from left to right) then from the engine's point of view the air moves from right to left, which is in the other direction: -4m/s. Likewise, it makes no sense for the exhaust velocity to be greater than the aircraft's velocity. Here I've used plain positive/negative values to show directions in a 1d space (since that's all we care about), but using vectors to indicate direction should work too.

I believe I didn't make any direction-related mistake.

If the inlet mass flow is 0 the engine is not running. You seem to be confusing mass flow rate with a velocity.

The thrust is same in both cases; the anomaly in your calculations is due to the fact that you are not accounting for the engine's own velocity.
In the first case, you say that the inlet mass flow rate is 0 kg/s because the air velocity at the inlet is 0 m/s. However, since the engine itself is moving forward at 4 m/s, the actual mass flow rate through the inlet is 4 kg/s, same as in case two.
Also, the exhaust mass flow shouldn't be greater than the inlet mass flow - that would be against the law of conservation of mass (unless the engine is running a ludicrously rich mixture). Since the exhaust velocity is greater than inlet velocity, the exhaust area must be smaller than inlet area such that the mass flow throughout the engine remains the same.
So the thrust in both cases is:
$$ F \  = \ ma \ = \ \frac{m}{t} × ∆v $$
$$ F = 4×5 $$
$$ F = 20 N $$

In both cases, thrust can't be the same because you need to look at momentum change of the air. In the engine-based reference frame, inlet and outlet velocities are much higher rather than in the ground-based reference frame. I think you didn't understand complexity of the question.

@Jawel7 I think you didn't understand the simplicity of the question :) In both cases, ∆v is 5 m/s. Also in both cases, the mass flow through the inlet (and therefore the engine) is 4 kg/s (explained in the 2nd para of my answer). So why should the change in momentum be different in both cases?

Who said inlet and outlet mass flow rate are the same? In no jet engine, they are the same because there's fuel addition.

@Jawel7 It looks like you didn't bother to read my answer.

But why are we using mass flow rate at inlet and outlet separately for thrust calculations in jet engines then? We are not assuming mass flow rate at the inlet and exhaust as the same? Can you make the same calculation with fuel injection? Let's say mass flow rate at the inlet is 1 kg/s and fuel is 1 kg/s and exhaust flow rate is 2 kg/s. Can you make the same calculation and find the same thrust for both cases? I couldn't find actually.

@Jawel7 When I said "unless the engine is running a ludicrously rich mixture", this is exactly the kind of mixture I was talking about. What kind of jet engine runs an AFR of 1:1? even pure nitromethane, which contains its own oxygen, has a stoichiometric AFR is 1.7:1. The kerosene used in jet engines has a stoichiometric AFR of 15.6:1, but jet engines run so lean that the actual AFR is closer to 75:1. Therefore, for all practical and purposes, the mass of fuel can be ignored, unless you specifically mention that the engine is actually running that rich.

Suppose the engine is afterall running that rich. So in the first case of the question the engine is burning 5 parts of fuel for zero parts of air. That's generally typical more for a rocket engine than a jet engine, but let's assume that that's the case. But then why is the air mass flow at the intake different for case two? The amount of air entering the intake is always the same, regardless of the frame of reference. In case 1, the intake is moving and the air is stationary, whereas in case 2, the air is moving into the stationary intake.

but in both reference frames, the amount of air passing through that 1 m² of area is the same. Now, you have to decide which of the two values for the intake mass flow is correct, because both of them cannot exist simultaneously. Once decided, we can determine the ∆v for both fuel and air seperately, and from that we can determine the thrust produced by the engine.

I think there's a mistake in the first case of the "observer on the ground" when it is stated that the "Inlet mass flow rate is 0 kg/s": that would imply that the engine is not ingesting any air i.e. that the inlet is simply closed and the engine not working.


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Why do we consider that air comes into jet engine?
While calculating air pressure and temperature during cruise of jet engines, we consider that air comes into the jet engine inlet with our flight velocity.
But why? In fact, air stops stagnant in atmosphere and aircraft moves with a flight velocity into air.
During calculations, we say: "Static pressure of air is , and it has a velocity of flight velocity. Then air comes to rest, so its static pressure increases!".
But the thing I don't understand is actually air doesn't have that velocity, what has that velocity is jet engine/aircraft.
Can you explain this assumption?

Velocity depends on frame of reference. If you were in the vacuum of space, and an asteroid was flying towards you, from the asteroids perspective it is static and you are flying towards it. Same thing is happening between the plane and the air, you're just using the nearby reference point of the earth to confuse yourself.

For a turbo-engine what ultimately counts is not the proper velocity of air, nor the velocity of the aircraft, but air velocity relatively to compressor rotor blades (or fan blades), which spin, that makes a big difference. Note when tests are made in  wind tunnels, this is air which is moving, not the surfaces.

Both the air and the engine are moving very, very fast. Our galaxy and everything in it is moving at a blistering speed of several hundred miles per second.
Of course, we don't feel this speed. Velocity is relative. Try sticking your hand out the window while driving (or flying, if you are brave and your POH permits it). The air pushes against your hand exactly the same as it would if you were sitting in a wind tunnel and your car wasn't moving at all.
This is the only reason wind tunnels are useful to begin with- because velocity is relative, the physics is the same whether the plane is flying through the air or whether it is sitting in a wind tunnel with the air being blown over it.

"Our galaxy and everything in it is moving at a blistering speed of several hundred miles per second." moving relatively to what? The Universe expands, but there is no center of expansion.

Actually, if the observable universe is in consideration, I am the center of it.

@mins It's the peculiar velocity of the Milky Way. Every galaxy would have a particular velocity if the expansion of the universe were perfectly isotropic, and the peculiar velocity is the difference between this velocity and the galaxy's actual velocity. You can also think of it as the velocity relative to the cosmic microwave background. This is the closest thing that the universe has to a rest frame and so the closest thing to an "absolute velocity" we can define. Of course, there is nothing actually privileged about this frame- the point of my answer is that only relative velocities matter

I originally had some of these details in my draft answer, but I think they detract from the actual point.

@Chris: In my hubble opinion, you're correct.

How is the physics the same? By taking the engine as reference frame, we are assuming air that is actually stagnant has dynamic pressure, which increases total pressure of flow. After this, we are doing all calculations based on it, and automatically flow conditions at nozzle are different than ground-based calculations too. In this case, kinetic energy change of the flow is different in jet engine-based reference frame than in ground-based reference frame. What it means? It means we find engine work on flow different for two different reference frames.

@Jawel7 The general principle is called Galilean relativity. As you've noticed, work done is not a Galilean invariant- the computation of work varies from reference frame to reference frame. The amount of work required to accelerate a mass a specific amount varies between reference frames by the same amount, so this cancels out, and no quantity that matters physically differs between the reference frames. The impulse on the air, and thus the engine's thrust, is a Galilean invariant and you will compute the same result in any frame.

@Chris-RegenerateResponse for thrust point of view, you are definitely right. However, we get the same understanding for energy of fluid as well, as I explained in my previous comment. So in this case, don't we get wrong values for energy point of view by getting the engine as reference frame instead of ground? Kinetic energy difference of the fluid, work done on the fluid by the engine, differs from frame to frame.

@Jawel7 Kinetic energy varies as the square of the speed, so the difference in energy of the fluid is indeed different from reference frame to reference frame. Not every reference frame agrees on the value of every variable. But you can do the analysis in any frame you like, and then use a Galilean transform to figure out what variables would be measured in any other reference frame without doing the whole analysis over. Conveniently, force is Galilean invariant, so every frame agrees on the actual number. Kinetic energy is not, so if you want to know that in another frame you need to ...

find it in one frame and then apply a Galilean transform to find it in another frame. It's not "wrong," it's just "different." Just like airspeed and groundspeed are different- they are just the speed measured in different reference frames, both are equally valid even though the number is different.

Because it doesn't affect the math, and because it helps with keeping relative change in perspective if we use the engine as a frame of reference instead of the airmass.
Also because you are incorrect in assuming that air "stops stagnant in the atmosphere". The air is in constant motion and has velocity relative to the earth.  However, if we used a fixed earth reference then we'd have to account for wind in our calculations and that would complicate things unnecessarily. (because where do you stop?  i.e. what about planetary motion?!)
We all understand that outside of a wind tunnel it is actually the engine that is moving, but it just makes a lot more sense to keep things simple and consider the engine as a stationary reference for doing things like describing the function of an engine and calculating airflow.

Whether the engine is moving is a matter of choice, both inside and outside the wind tunnel. Motion is relative and no reference frame is better than any other (well, reference frames in free fall are a bit better, but on the Earth those don't tend to be co-moving with anything useful).

@JanHudec, I disagree completely that "no reference frame is better than any other" for the reasons given in my answer.  Consider the verbal gymnastics needed if you used ground track with a stiff crosswind as your reference when explaining the path of airflow through a centrifugal flow compressor.  Why not eliminate unneeded variables?

Well, we have to distinguish between better for describing particular situation and “always” better. The former definitely exists (and is the reference frame attached to the engine in this particular case, which is the point here). The later does not, that's the whole point of relativity.

I didn’t use the term “always”.  I’m generally pretty careful to avoid absolute statements unless they are “always” true.  Your statement “no reference frame is better than any other” is pretty absolute, despite the exception in parenthesis.  I’m always open to constructive feedback for improving my answers, but I don’t see anything helpful I can actually use in this.  Unless you can offer a more specific suggestion?

I quite like the answer as written (and upvoted it before I even wrote that comment). I only commented—and admit I was very unclear—the bit that the engine ‘is actually moving’. You can always choose a reference frame in which they are moving, but it isn't actual, it's still just a choice, and it would be a bad choice here.

Jan - I'm afraid I have to agree your comments (in this case) came off a bit muddled!    In your sentence " ... is better than any other  .." you're referring to what we all learned in junior high (primary, out side the US  :) :) :) ) which has no connection to this answer's use of "better".  BTW, note too that "Whether the engine is moving is a matter of choice, both inside and outside the wind tunnel" is very confusing, it's incredibly hard / impossible to make an engine in a wind tunnel move, ie have any velocity (ie, compared to the building) for more than split-seconds.

@MichaelHall: In this context, "better" is (probably) not being used to mean "more useful," but rather to mean "more correct." And it is absolutely true that no reference frame is more correct than any other reference frame. That probably seems like an obvious truism to you, but it can be a bit of a revelation for physics students who have not yet encountered it.

The answer to the last question is "Galilean relativity", the assumption that the same laws of physics hold in any inertial frame of reference. In the example, it is presumably easier to do the analysis in the engine's frame of reference.

Welcome, you win for succinct!

Our universe is so constructed that it doesn't matter if the air is moving and the model is standing still, or if the model is moving and the air is standing still. In either case, air is coming into the jet engine.
Note also that when a jet-powered plane is operating at its design point for cruise conditions, the air right inside the engine inlet is moving at the same speed as the air just outside the inlet. That is, the streamlines of air parcels entering the engine are not perturbed by the engine cowling. There is no spillage nor suction effects at design point cruise.

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Split Flaps and lift?
Do split flaps produce lift? I don't see how, because there is no change in camber.
It seems like an upside down speed break, producing only drag

Yes, deploying split flaps increases lift, because they do increase camber.  It's as if the airfoil became solid between its (now two) trailing edges.  Between those two trailing edges there's a lot of turbulence, of course, but that doesn't negate the fact that camber increased.

A split flap is a very good choice for the landing configuration. Simple and effective.
For the approach you want that extra drag. It helps to slow down and allows to keep the engine running at a higher thrust level, so spoolup will be faster once you decide to go around (and raise the flaps, hopefully).
Split flaps do increase lift and are even more effective for the same chord and deflection angle than a simple camber flap!
In the words of Sighard Hoerner (see section 5.7 in his book Fluid Dynamic Lift):

Figure 7 reveals, however, that split flaps are somewhat superior to plain flaps. The straight upper side of the airfoil section evidently leads the flow right to the trailing edge, while past a plain flap, the flow is exposed to interference (mixing) with the highly turbulent "dead" space. Another advantage of the split type might also be a more stable wake pattern; the separation points are clearly fixed.

And these are Figures 6 and 7.
If you want even more lift for a bit of added complexity, make the hinge point move back when the flap is lowered. This is the Zap flap, named after Eduard F. Zápařka, who invented it in 1931. Moving the flap back in effect increases wing chord. It is not only the Fowler flap (which was invented by Gustav Lachmann, btw.) which performs that trick.

Figure 8, from the same book and chapter.

Do split flaps produce lift? I don't see how

Well, after all lift is created by deflecting air downward, which is exactly what a split flap does - although in a very inefficient way i.e. with a lot of drag.
This NACA TN shows how $C_l$ and $C_d$ increase with the deflection $\delta_f$ of the split flap (left plot):

At stall (at around 10° AoA) $C_l$ goes from 1.3 to 2.2 (plus 70%)  while $C_d$ goes from 0.1 to more than 0.4 (plus 300%!).
For comparison, the following plots show the behaviour for a double slotted flap, like seen on almost every jetliner (source this NASA TN):

With this more efficient flap configuration, to get the same increase of 70% $C_l$ at stall, the flap has to be deflected of 35° ($C_l$ increases from 1.6 to 2.9); anyway $C_d$ remains almost the same at 0.026!

It seems like an upside down speed break, producing only drag

A speed brake does not increase drag only but decreases lift as well.

These hard numbers are more convincing than my handwaving :-)

@CamilleGoudeseune Nevertheless, your intuitive picture you drew was really helpful to me.  I like reading answers that come from different directions.

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Is Centre of Gravity located behind or in front of Aerodynamic Centre?

I was reading CAE Oxford POF book 2014 edition found online, and in lesson (10-Stability and control), page239, it stated that CG must be BEHIND AC in order to maintain aircraft's stability, but I found a lot of videos and articles online stating the complete opposite, that the CG must be IN FRONT of AC. Now I don't know which is the correct answer, can anyone confirm which is true, and if so, why is the other opinion is making up this statement

This question is similar to: Why do most commercial aircraft have the Center of Gravity before the Aerodynamic Center?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem.

Can you post the link(s) of the online article(s) you found stating the opposite?

@sophit This appears to be one of the versions of the book. It's on page 260 (so different than OP's 239). The site requires you to enter page 266 to arrive there. The image is "Figure 10.19 Unstable wing contribution", so it appears OP misunderstood it as being the stable configuration instead.

@levantpied: thanks for the link that I read. What CAE affirms seems to be indeed correct. It would be nice to read the other sources to make a comparison.

https://online.pubhtml5.com/aafi/cxah/#p=267 here is the book where i got this info sophit, even in the pages before this one, we can still see the CG is behind the AC

For static stability, the "aerodynamic center" refers to the entire aerodynamic footprint of the aircraft, wings, tail, nacelles, fuselage, that must be behind the C of G. You could say that it's a weathervane effect.
If you want the weathervane to point into the wind, the point at which the forces of the wind are focused has to be behind the pivot. The point of focus is the aircraft's aerodynamic center when factoring in all the various moments acting in the pitch axis, also known as Neutral Point.
In the case of the aircraft, as a free body in space, the pivot is the C of G. Think of it as a kind of a free floating weathervane turned on its side, that can be trimmed to align itself at an off angle to the airflow to get lift from the wings by forcing them into a positive angle to the flow (without the trimming force, it becomes, effectively, a lawn dart, and will arc into the ground with perfect static stability).
If the C of G is on the wrong side of the aerodynamic center, then all you've done is turn the weathervane backwards and it will want to switch around to point the other way.

The aerodynamic center of the wing is where it has no tendency either to rotate away from or back into the wind regardless of pitch.
But let us not forget the fuselage and tail, which tends to move the aerodynamic center of the entire aircraft rearwards.
In order to have longitudinal stability, the CG must be ahead of the aerodynamic center.
Next comes the relationship of center of pressure (or lift) of the wing and the CG.  In your diagram, they could be closer together and/or have a matching torque contribution from the tail.


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Is washout a twist up at the root or a twist down at the tip of the wing?
Does washout/in take place at the root of the wing or the tip? And is the twist distributed evenly along the wing, or is it only twisted for a portion and level for the rest, specifically in commercial airliners?

"Does washout/in take place at the root of the wing or the tip?"  Have you heard the term "frame of reference"?

Washout is chordline twisting nose down toward the tip. When it starts is not important. There is washout if the angle of incidence of the tip is lower than the root.
Wash-in is the opposite.
Sometimes the effect of washout is achieved by changing the airfoil contour toward the tip without actually twisting the wing, with the change to an airfoil with a higher stalling AOA near the tip. Or you might achieve this simply by adding a leading edge cuff or droop to the outboard end of the wing which does the same thing.


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Are flap track fairings (just the appendages themselves) necessary for wing aerodynamics?
Coming from this answer by Peter Kämpf leaving me a little perplexed about the necessity of flap track fairings. without flap tracks, would we still need them on the wing of any given sub/transonic aircraft?

"without flap tracks, would we still need them on the wing", it's not clear what is "them", do you mean: Are flap track fairings required for sub/transonic aircraft?

@mins sorry, "them" was referring to the fairings.

This might also help.

No.  The use of flap track fairings as a way of influencing transonic aerodynamics is called Kuchemann Carrots.  Because they kinda look like Carrots and they were proposed by Dietrich Kuchemann.
A modern clean wing (without the antishock bodies) could be designed using a different airfoil and perhaps different planform.
Of course, modern clean wings do need flap tracks (to hold the flap deployment mechanisms and structure) and so some characteristics of this sort of design still persists.  You want to make the flap tracks as small as possible -- but then shape them to interact as favorably as possible.

You do your best to incorporate the tracks into the airfoil profile so as little as possible is sticking into the airstream when they are retracted. To the extent you can't do that and you have bits protruding, you fair the tracks like any appendage or mechanism sticking out into the airstream and adding parasite drag.
You may elect to do this and live with fairings (sometimes called "canoe fairings" on airplanes with long-travel Fowler flaps whose tracks project below the trailing edge) because structural or mechanical design priorities make it worthwhile


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What type of airfoil should be used for horizontal tailplanes?
Don't aircraft use symmetrical tailplanes so as to not produce lifting forces when not needed? Im confused by people saying that they use upside-down cambered airfoils.

Why are you confused? Do you think cambered airfoils have no zero-lift angle of attack? (assuming zero-lift is required in horizontal flight, which is not the case)

What do you mean with "upside-down cambered airfoil"? Do you mean reflexed airfoil?

Very related: Does the tailplane have a positive/negative camber?

@sophit My fault, I meant a cambered airfoil turned upside down.

@mins Okay, I'm probably stupid, but wouldn't that angle be totally perpendicular? or otherwise at stalling level?

There are certainly no stupid questions, I'm trying to understand what does confuse you. On this cambered airfoil an AoA of -5° creates no lift (lift coefficient is 0), at more than -5° it creates lift, more and more, until it starts stalling between 15° and 19°.

Most statically stable aircraft with the tail aft have a download on the tail.  The most efficient way to do that is with a horizontal tail with a cambered airfoil -- placed upside down.  This configuration is quite common.
You will sometimes see symmetrical airfoils used, but this is not because they run at zero load.  These horizontal tails are typically placed at negative incidence relative to the wing (the relative incidence between the wing and the horizontal tail is called declage, positive when the wing is at larger leading edge up angle).
I do not know of any aircraft whose vertical tail is not a symmetrical airfoil.  Some vertical tails (on propeller aircraft) are mounted at an angle to build in some offset for p-effects.
Edited to correct that negative tail download is not absolute.  Thanks Peter Kämpf.

The Me-109 had a cambered vertical tail. When MBB built a 109 G from parts of a Spanish HA-1112, they initially used a symmetrical airfoil on the tail, only to later find out how the original tail looked. When the old symmetrical tail had been switched for the cambered one, takeoff characteristics greatly improved.

"Statically stable aircraft with the tail aft always have a download on the tail" - no, that is not always true. Many gliders have positive tail lift at low speed.

@PeterKämpf Interesting. Won’t any cambered airfoil create a torque in which the H-stab needs to provide downforce to counteract it? (Probably depends on if the wing’s airfoil has camber)

@Wyatt Like mounting the vertical tail at an angle, I'm sure using a cambered vertical tail was done to offset P-effects.  P-effects is a generic term for aerodynamic effects of the propeller slipstream on the aircraft.  Not only does the propeller increase the velocity of the streamtube, it also adds a significant swirl component to the velocity.  That swirl hits the lifting surfaces as a local change in angle of attack that can cause the aircraft to yaw or roll.

@RobMcDonald I see. I also wonder if adding contra-rotating propellers could minimize the p factor issue. As for Peter Kämpf’s other comment, won’t any cambered airfoil create a torque which the H-stab would need to make downforce to counteract? (It might depend on CG location for one thing)

@Wyatt  A counter rotating center-line prop would.  Likewise, many multi-engine propeller aircraft use counter rotating propellers.  The effect of camber or incidence of the vertical tail on the load distribution of the horizontal tail will be very small -- but it is added in to try to reach neutral behavior (when including P-effects).  So we don't really worry about breaking it all down, but adjustments seek neutral overall behavior.

I think your doubts arise from confusing the aerodynamic characteristics of a cambered airfoil versus an uncambered aka symmetrical airfoil.
The following plot shows the lift coefficient for a symmetrical and a cambered airfoil i.e. the classical NACA 0012 and NACA 4412 respectively (source):

The main aerodynamic effect given by cambering an airfoil is basically to shift the lift curve upward. Comparing the previous two plots you can see how the $C_{l_{max}}$ increases from some 1.3@17° for the NACA 0012 to 1.5@15° for the NACA 4412 while the plot intercept the x-axis at 0° for the symmetrical airfoil (obviously) and at some negative value for the cambered one (-3°).
(Even if it is not shown, the plot for the symmetrical airfoil continues - symmetrically - at negative AoA reaching -1.3@-17° while the NACA 4412 should reach -1@-19°).
So the main reason to camber an airfoil is to achieve higher $C_{l_{max}}$ @ lower AoA.

What type of airfoil should be used for horizontal tailplanes?

The main purposes of the horizontal tailplane is to trim, to manoeuvre and to give stability around the longitudinal axis of the airplane.
Being on the tail is a sufficient condition to provide the airplane with stability, no matter if the airfoil is symmetrical or cambered. Conversely, to trim and to manoeuvre the airplane both a positive and a negative lift is normally needed with a zero (or even slightly positive) lift being the ideal condition: this implies that in general a simple symmetrical airfoil like a NACA 0012 or 0009 is a good choice for any horizontal tailplane.


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2D vs 3D Pressure Drag
What is the difference between the parallel component of the drag that depends on alpha for a 2D airfoil and the parallel component of the drag that's on a finite wing?
I understand that for a wing the spanwise flow causes a loss in lift, is this also what causes a parallel component of the pressure differential along the streamwise direction to develop?
The reason I ask is because I've heard people explain that downwash causes this additional pressure drag for 3D wings, but downwash does exist for airfoils too, yet any alpha-dependant drag is bookkept as form drag for 2D airfoils.
I tried avoiding the use of the term, lift-induced drag or the phrase tilting back of the lift vector.


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Why are inlet and outlet (exhaust) pressure considered to be the same in a turbojet engine?
Why do we take outlet/exhaust pressure (station 6) as the same as in inlet (station 1) ?
In T-S diagram, we see that outlet temperature is much higher than inlet temperature since flow in outlet has higher kinetic energy. It is hot and has more velocity than inlet.
So I know that static pressure of inlet and outlet is the same, but we use stagnation pressure/temperature in station 1 and 6, so I think that stagnation pressure of station 6 is higher than 1 because it has more velocity and temperature.
What do you think?

What do you think the pressures at #1 and #6 are?  Why would they be different from ambient?

EPR, or Engine Pressure Ratio, is an interesting subject.  It seems, in reality, EPR > 1 as more throttle is applied.  In theory, if P6 > P1, V6 may be less than optimal.  But jets seem to like EPR more around 1.2 - 1.4.  May have something to do with less flow turbulence around the nozzle exit.  Learning more about it at present.

My memory on this is a little foggy -- but I think your observation is correct.
When doing cycle analysis, I think we apply a static pressure condition to the nozzle exit (not stagnation).
I think it is drawn that way in the cycle diagram because eventually the air returns to ambient conditions (after lots of mixing and heat transfer).  The dotted line is basically a part of the overall cycle that we don't model in detail.  I wouldn't assume it follows a strict constant stagnation pressure line.

Well, the trick to keeping that thing running is to keep everything moving.
Piston engines are batch thrust producers, and jets are continuous thrust producers.

Piston engines produce thrust?

What would you call a linear force in a given direction?  How about this:
$\vec{F}$.  On the pistons.  Which turn the crankshaft.

Piston thrust over time produces an average force.

Look at a radial engine.  Look at 4 of them bolted together.
The mass of the moving engine parts and the prop also help smooth things out by acting as a momentum conserving flywheel.

because piston engines are used to create circular mechanical motion, their Force output is measured in torque, or pound-feet.

Or ... Horsepower as pound-feet/second.

all by expanding the volume of gas by heating it.

Pressure × Volume = Gas Density x Temperature

As you heat gas either the pressure or the volume will increase.  After spark and combustion, increased pressure in the cylinder creates the FORCE that moves on the piston head until cylinder pressure is closer to incoming pressure.

the result is higher VOLUME of hotter gas

Then pushed out into the exhaust, it's work done.
Now on to jets.  They, as everyone knows create thrust ... and their thrust is also unaffected by velocity ... and they also create mechanical force by expanding gasses ... which turn a turbine ... which can (for turboprops) be measured as pound-feet or torque per second!

so what is it about jets?

The flow in and out must be continous or the jet will not run properly.  While piston engines make thrust in little batches$^2$ within a closed cylinder with variable pressure, jets produce a continous "push" on the turbine with constant pressure.
This is a very delicate balancing act, and a real challenge for early jet designers.  If incoming and exiting pressures are out of balance, the  compresser can "surge" or stall and the continous thrust is lost.
This is why computers, such as FADEC, have greatly reduced the frequency of these events occurring.
Theoretically, the faster exit gasses should be of lower pressure, and indeed, pressure reduces passing through the energy extracting turbines, but the jet engine operator is most concerned with keeping flow rate through engine, and pressure in the engine, steady by adjusting intake and exhaust volume and temperature.

Pressure = Density × Temperature/Volume

As in piston engines, the exhaust pressure may be slightly higher than inlet.  And like a piston near the end of its power stroke, the pressure difference is so little that:

it isn't worth the weight of an extra turbine to try to extract any more mechanical energy

So it goes out the back as ... thrust!$^1$
$^1$ design depends on the airspeed.  At lower airspeeds, turboprops (and helicopter rotors) are far more efficient than turbojets, so instead of just making thrust as (m$_{\text{2}}$V$_{\text{e}}$) - (m$_{\text{1}}$V$_{\text{o}}$), the mechanical energy from the turbine is used to turn a rotating airfoil.
$^2$ these "little batches" of thrust produce a "pulse" in exhaust pressure, which can be "tuned" to increase efficiency of 2 or 4 cycle engines.


text:
Why do we take into consideration holes for pressure thrust in jet engine calculations?
In many sources related to jet engine thrust calculations such as https://www.jet-x.org/a8.html ,
pressure thrust of a module such as combustion chamber, nozzle or compressor, is calculated by front and rear cross sectional area of them (including holes at their center) and flow pressure at front and rear of them.
If they were a complete solid disc without any hole, I could understand this calculation.
However, for turbine in an example, it has a hole at rear and front, so either pressure at rear can't act on the hole at rear or pressure at front can't on the hole at front. In my opinion, these holes should be taken out of cross sectional area term.
Why do we think as if fluid pressure is able to act on the whole cross sectional area including holes at the center whereas they can't physically?

Can you be more specific? The linked source is very long. Which portion do you think is wrong? And what do you think the correct answer should be? E.g. In section x they use use area Y m^2 but you think it should be area Z

A diagram might help understand your question.

Specifically, where do you see the page state the cross-sectional area includes the “holes” (which are in fact opposite of holes, because it's the body with the shaft)? I only see them say “nozzle cross-sectional area” and that is just the area of the nozzle itself (which has annular shape).

This answer assumes that by "hole" you are talking about a non-physical boundary like the exit plane. If so, it's explained here:
https://www.grc.nasa.gov/WWW/k-12/airplane/thrsteq.html

There is an additional effect which we must account for if the exit pressure p is different from the free stream pressure. The fluid pressure is related to the momentum of the gas molecules and acts perpendicular to any boundary which we impose. If there is a net change of pressure in the flow there is an additional change in momentum. Across the exit area we may encounter an additional force term equal to the exit area Ae times the exit pressure minus the free stream pressure. The general thrust equation is then given by:
F = (m dot * V)e - (m dot * V)0 + (pe - p0) * Ae
Normally, the magnitude of the pressure-area term is small relative to the m dot-V terms.


text:
Do compressors and turbines in jet engines end with rotor or stator?
I am trying to understand compressor and turbine design principle based on stators and rotors.
I know that turbines and compressors both have a rotating and fixed section that constitute a stage.
For compressors;
1-) Do they begin with rotor and end with stator? (I think there should be rotor at first to capture air in precisely and there should be stator at the end of compressor and in the beginning of combustion chamber since combustion chamber needs the highest pressure of fluid, so we can maximize fluid pressure by slowing down with stator)
For turbines;
2-) Do they begin with rotor and end with rotor? (I think there should be rotor in the beginning of turbines and at the end of combustion chamber to capture air in precisely and there should be rotor at the end of turbines as well since we need dynamic energy instead of static pressure, so we can maximize dynamic energy with rotor just before nozzle.)
Are my understanding and information correct?

Having a look inside is always a good way to learn. Let us know if something isn't clear.

@sophit I looked at it before actually, but am not sure about answer to my question. It seems that compressor begins and ends with rotor, turbine begins with stator and ends with rotor. Am I correct? I think I am not. Because I think if they begin with rotor, they need to end with stator, or exact reverse.

Well, if you consider the fan as the very first stage of the axial compressor then it does begin with a rotor. Just before and after the combustion chamber there are indeed stators to guide the airflow in and out. After the turbine(s) there might be some guide vanes. But let's see what the engine experts say.

@sophit Older low-bypass engines, like the JT8D, had inlet guide vanes before the fan, so those would be stator. I think the answer to this is "it depends".

@user71659: that's why I wrote only a comment. But it looks like you have the answer

There is a possibility that it starts with a rotor and ends with a rotor (contra-rotating spools).

A compressor stage is rotor followed by stator. Many, but not all, have inlet guide vanes (IGV)at the inlet.
Turbine stages are vane (stator) followed by blade (I had a board designer not even know the term, to him the rotor is the spinning shaft). There can be exit guide vanes to straighten the flow.
The IGV forces a tangential velocity on the flow, and the axial component cannot change much (the same flow ahead and behind the IGV) so it does what the turbine vane does: increase speed and decrease static pressure. That gives less problem with separation. Exit guide vanes decrease the tangential velocity and therefore velocity so they work like compressor stators.


text:
Why does pressure difference between top and bottom surface of wings increase as speed increases?
We observe that lift force increases as speed increases on aircraft wings. Theoretically, there must be increase in pressure difference between top and bottom surface of wings.
What I don't understand is why pressure difference increases as speed increases?
Air velocity increases on top of wings, but not only top of wings, it increases on bottom of wings too. So pressure on top decreases, but pressure on bottom decreases too. Why pressure on bottom decreases less than it does on top?

In this link, he proves lift equation but in 4.23, he removes delta V(V2-V1) as writes it as only V.
Why he converted V2-V1 to only V ?

EDIT:
Just realised that there's another big mistake in that video i.e. the "equal transit time" hypothesis. The higher speed on the upper surface in respect to the lower one is justified saying that air travels a longer distance on the upper surface and a shorter distance on the lower surface and afterwards it rejoins behind the airfoil... that's just plain wrong as explained for example here.

he removes delta V(V2-V1) as writes it as only V. Why he converted V2-V1 to only V?

Because he's making a very huge and basic mathematical mistake! At  minute 3:50 he substitutes $(V_2²-V_1²)$ with $∆V²$ but by a mathematical point of view the latter correctly expands to:
$∆V²=(V_2-V_1)²=V_2²+V_1²-2V_2 V_1$
which is definitely not equal to $(V_2²-V_1²)$!

Air velocity... increases on bottom of wings too

That's another mistake! Normally the opposite is actually observed, i.e.:

an increase of speed on the upper surface and;
a decrease of speed on the lower surface.

Due to the momentum conservation (aka Newton's second law aka Bernoulli equation), an increase in speed is "compensated for" a decrease in pressure and viceversa. That implies that we get:

a lower than ambient pressure acting on the upper surface and;
a higher than ambient pressure on the lower side.

The upper surface is therefore "sucked upward" and the bottom surface is "pushed upward". The sum of these two effects is an aerodynamic force mainly directed upward (obviously).
Just as an example, the following picture taken from this paper shows the typical distribution of variation of pressure along the chord of typical airfoil (NACA 0012 @ 8° AoA):

The upper curve is the plot for the upper surface and the lower curve is the plot for the bottom surface. Negative numbers mean that the local pressure is lower (but obviously not negative) than ambient pressure and viceversa.

According to Bernoulli equation, it's important V2^2 - V1^2. Why V2 (velocity on top) increases much more than V1 (on bottom) does?

There is no suction. There is only pressure. "Suction" is just a word we use to talk about positive pressures that are less than normal ambient atmospheric pressure, (14.7 psi),

Even in this answer, the quote states that "Negative numbers means lower than ambient pressure and vice versa." The actual numbers are NOT negative. They are just less than 14.7 psi.

The plot has both positive and negative numbers on the y-axis, no idea what you're talking about...

@Jawel7 For a given angle of attack, shape of airfoil and sufficiently low Mach numbers, the pressure distribution described by the chart that this answer shows is always the same. There is not really a "why" to it, that's just how air in the physical world behaves. We know that pressure is proportional to the square of velocity (again, that's just how it is), so you can see from the chart that the square of the velocity at the bottom of the airfoil is roughly 0.8 times the square of the freestream velocity (cp = 1 - u^2/u_inf^2 = 0.2, so u^2/u_inf^2 = 0.8) ...

... so the velocity at the bottom of the airfoil will increase with increasing freestream velocity, if you keep the angle of attack constant. It will always be about 0.9 (square root of 0.8) times the freestream velocity.

Because lift is not suction. "Suction" is just a way we describe a pressure that is less than some defined "standard" atmospheric Pressure. The air flow on the top of the wing is still pushing down on the top with positive pressure. It's just less than the pressure pushing up on the bottom.
So as velocity increases, the pressure increases on top and bottom equivalently. As we increase velocity, the pressure goes up by the same percentage on both top and bottom. But if it's 100psi on the top, and 105psi on the bottom, and both increase by 10%, then it becomes 110psi on the top, and 115.5psi on the bottom. The difference has also increased by 10%.

As seen in sophits answer, particularly for various portions of the upper and lower wing, there is indeed positive and negative pressure relative to freestream.  What should be reviewed is upper wing pressure increases with velocity.

Indeed, the upper wing vacuum pump becomes more effective, allowing lower pressures, as V increases, especially when Reynolds number exceeds around 500,000.

It is amazing how we continue to struggle to understand the fundamentals of lift, but no need to be tearing apart somebody else's work.
In the video (rightly or wrongly) the Bernoulli equation for flow through a restricted pipe is used:
P1 + 1/2 $\rho$v1$^2$ = P2 + 1/2 $\rho$v2$^2$
F/Area = (P1 - P2) or $\Delta$P
Where he get a little shakey is when a division by $\rho$ step is not explained, leading to
P1/$\rho$ + 1/2 v1$^2$ = P2/$\rho$ + 1/2 v2$^2$
Now we have $\Delta$P/$\rho$ = 1/2 (v2$^2$ - v1$^2$) = 1/2 $\Delta$ (of) v$^2$
Plugging back in to $\Delta$P/$\rho$ = F/(A$\rho$) = 1/2 $\Delta$v$^2$
--> F or Lift = 1/2 $\rho$ × A × $\Delta$v$^2$, or, substituting the symbol S for area :

Lift = 1/2 $\rho$ × S × $\Delta$v$^2$

Although the math is correct: the issue is that $\Delta$v$^2$ in pipes is further dissected to v$^2$ of the freestream × coefficient of lift in aviation.
Coefficient of lift contains the factors of Angle of Attack and Airfoil shape and camber.
Any object moving through a fluid will create drag, but we can study the phenomena relative to the aircraft, so now we have air flowing over the stationary object, as in a wind tunnel.
Perhaps easier to understand by taking a fully symmetrical object, and pointing it straight into the wind: we have high pressure and lower velocity in front, and lower pressure and higher velocity on top and bottom.
In order to create lift, we must have an angle of attack to the freestream.  Lift can be positive or negative, but to counter-act gravitational force, we need to produce a postive angle of attack$^1$.
Now, the higher pressure, lower velocity region grows in size and shifts to underneath the wing, and the lower pressure higher velocity "wake" of the wing is on top of the wing.
Naturally, we see an increase in drag as AoA increases.

But as speed increases, the ability of the wing to hold higher pressure below, and lower pressure above, increases because $\Delta$ pressure can be maintained faster than the natural tendency of air pressure to equalize pressure.

Here the energy of the thrust is literally running a vacuum pump and a pressure pump on the wing at the same time, creating the lifting force.
Another very important factor is Reynolds number.  Above a certain limit, the efficiency of the pressure/vacuum pump increases greatly, allowing much higher Lift to Drag ratios.
Changes in airfoil shapes design can improve this further.
$^1$ asymmetrical wings can generate lift at 0, or even slightly negative, Angle of Attack

My way: $\Delta$P = F/A = 1/2 $\rho$(v2$^2$ - v1$^2$).


text:
What causes the horn-like noise during the start-up of some jet engines?
Here's the specific sound I'm talking about (it starts specifically 56 seconds into the video, as well as at 2:02 for the second engine):

I've only really heard that sound on low-bypass engines (F/A-18s, BAE Hawks, etc); they don't occur right at engine-start (so as far as I'm aware it's not just bleed air being blown into a turbine for spin-up); I was thinking that it's perhaps either fuel flow causing an increase in exhaust flow through a slowly-moving turbine, or that it might be the hydraulic pump (or something similar) connected to the engine accessory drive.

You can hear that on high bypass engines, too. I believe this is related to flame stabilization in the burn chamber at low speeds. Combustion is not stationary causing pressure fluctuations and thus the sound

Almost all jet engines will have one or more "resonance" conditions where the combination of turbine RPM, mechanical elasticity and aeroelasticity constructively reinforces to cause mechanical or aerodynamic vibration.
Think of it this way:  If you have a slinky spring (mechanically elastic) toy you can hold it by one end and let the rest drop freely below that.  Now if you move your hand up and down at different frequencies (up and down and back up again in 5 seconds, 4 seconds, 3 seconds, 2 seconds, 1 second, 1/2 second) you will find that there is one rate that will cause the most deflection/movement of the spring toy even though you are moving your hand through exactly the same distance. This is the one closest to its resonance.
Now the ductwork and spools of a gas turbine engine are just a little more complicated than our Slinky toy, but still a mechanical (in this case, aero-mechanical) system susceptible to resonance conditions.  The trick is to design the system so that these resonance conditions, which can be mechanically destructive, don't happen often or for very long.  As the turbine RPM increases through the resonance range things start to vibrate (in this case at ~250-275 Hz) wildly enough to create that extra sound.
In fact, the designers of that GE F414 engine could tell you exactly the RPM range where that happens.  The aviators and groundcrew are attune to the sounds of this machine and would likely shut things down and have a closer look if this particular phenomenon didn't happen as expected.
So, the horn-like noise is vibration of the mechanical parts of the engine ductwork and the air flowing through it for the short period of time that the engine RPM and airflow align to make this condition happen.


text:
How does one choose the number of blades and the diameter required for a turboprop propeller?
I am interested in the preliminary design of a regional aircraft that uses turboprops as the propulsion method.
Following considerable research online about propeller design (disk theory, BEM...) I am rather lost on where to start.
To put it short, what parameters must be considered when designing the propeller of a turboprop device?
I am mainly interested in obtaining an idea of how many blades would be required and the diameter of the prop.
I have obtained values for the engine power requirements, and from what I can gather the main reason to have an increased number of blades is due to the requirement of absorbing the power from the engine. The latter therefore implicating an increased solidity ratio.
Is the any standard formula/theory that can provide such answers?
Thanks,
FZ

The basic rules are:

Make the diameter as large as practical.
Now you will most likely run into problems with supersonic tips. Reduce RPM and diameter until the tips are below Mach 1.
Now you will most likely not be able to absorb the maximum power of the engine. Increase blade count and chord until you do.

In the end, there is no simple formula because you need to find a compromise between conflicting demands. Increasing the number of blades will make the hub more complex and reduce propeller efficiency. Increasing chord will increase centrifugal loads and friction drag.
Try to find existing designs with the same disk loading. Your propeller should look very similar - existing designs went through the same design process.

The prop diameter is chosen to provide suitable ground clearance for the desired installation. Then the reduction gearbox ratio is chosen to keep the prop tips subsonic. Then the number of prop blades is chosen to absorb full engine power. Compromises then get made to take advantage of off-the-shelf props in the market and standard gear ratios offered by the engine manufacturer.

Thank you for your answer. Is there any theory that you know of that determines the number of required blades to absorb the full engine power?


text:
Why doesn’t the A320 family suffer from the same design constraints regarding engine placement/landing gear length as the Boeing 737 family?
This answer gives a pretty good description about the design issues which lead to the problems of the Boeing 737 Max (basically the same as in a video I watched a while ago):
The landing gear of the 737 is fairly short. One benefit at the time was that this eased baggage/cargo loading by hand. The JT8D engines for which the original 737 was designed were slim enough to fit under the wings.
However, these engines performed poorly when compared to later generations. Unfortunately, later generations of engines such as the CFM56 are also larger in diameter, so Boeing had to resort to various tricks to squeeze modern jet engines underneath the wing, while still ensuring sufficient ground clearance.
Increasing the height of the landing gear, thus adding enough ground clearance in order for the bigger engines to fit underneath the wings, was not an option. As the landing gear retracts into the fuselage in flight, its length essentially is limited to the distance between its mount points on the wings, minus some clearance.
Mounting the landing gear further outward on the wing would have required a structural redesign to cope with the increased torque from the landing gear. Retracting the landing gear with the wheels in “tandem” (one behind the other rather than both next to each other), or shortening the gear upon retraction (either by a telescope mechanism or an extra hinge) would also have been major design modifications, which would have been costly and taken extra time to develop.
Therefore, Boeing opted to reduce the engine diameter as far as they could (which helped a bit but not enough), and move the engine further forward and higher up on the wing. Because the latter alters certain handling characteristics, they added an electronic system known as MCAS to compensate for that. As this change was not clearly communicated, pilots were not aware of it, which eventually lead to two crashes and the grounding of the entire fleet.
What none of these two sources addresses: the 737’s main rival, the A320 family, was designed two decades later and had the larger engines from day one, along with sufficient space underneath the wings to accommodate them.
So how does the A320 design address this? Longer landing gear with attachment points further outward on the wing (which I assume is fairly easy to do on a new design but comes at a significant cost when upgrading an existing one)? Or something else?

So how does the A320 design address this? Longer landing gear with attachment points further outward on the wing [...]?

Yes, the A320 has a longer and therefore wider landing gear. You can find general aircraft dimensions in the Airplane Characteristics for Airport Planning documents.
Airbus A320-200: 7.59m wide landing gear

Boeing 737-800: 5.72m wide landing gear

The same document also lists the ground clearances at different points. At MTOW (max. takeoff weight) or MRW (max. ramp weight), the ground clearance
below the fuselage is
A320: 1.72m, 737-800: 1.30m
and below the engine nacelle it is
A320: 0.57m, 737-800: 0.48m

Sources: Airbus, Boeing

So, essentially, the A320 benefitted from being designed 20 years later and thus for more contemporary jet engines than the 737, and as a result, the design could be optimized for that. While the A320’s designers had an incentive for a long landing gear (more space for larger engines), the designers of the 737 might even have had an incentive for a short one (manual baggage/cargo loading).

@user149408 Exactly right. When the 737 was re-engined the first time (to the CFM56), they only barely got away with it by flattening the bottom of the nacelle. If they couldn't have done that, Boeing would probably have developed a new short- to mid-haul airliner replacing the 737 a long time ago. The A320 was designed for the CFM56 (and IAE V2500) with a normal round nacelle in the first place, so much higher up.


text:
Will the V280 operate with only one engine?
Can the newly contracted Bell V280 Valor operate on one engine?

Bloody well better or you won't get me in the thing.  The Osprey has cross linked transmissions, so I can't believe this one doesn't. Tilt rotors can technically autorotate, but you can't really land them in one piece from an autorotation, so you better hope a double engine failure never happens.

@John K cross linked transmissions suggests a single point of failure, although hopefully much more reliable than an engine.  Still, that’s not entirely reassuring.

According to Wikipedia: "A driveshaft runs through the straight wing, allowing both prop rotors to be driven by a single engine in the event of engine loss."

@Frog Virtually all helicopters have a "transmission", so that's a single point of failure. Helicopters also have the so called "Jesus nut" which is also a single point of failure.

@Aditya Sharma you’re right, a single point of failure is sometimes unavoidable but it’s never a desirable thing

The so called "Jesus nuts" doesn't exist anymore in any modern helicopter, only on Wikipedia. Actually the whole transmission+shaft+rotor head+blades is a single point of failure.

@Frog You overcome single-points-of-failure by over-building and robustness such that components are immune to fatigue, and/or have mechanical overdesign that result in obvious detectable degradation before something lets go. For ex you may design a drive shaft system with redundant bearings, and, say, a dual-wall shaft where a secondary load path takes over when one fails, AND a way to detect such failure when it happens to minimize dormancy. Stabilizer trim jacks on airliners do this.

@Frog I don't think this qualifies as a single point of failure. If the driveshaft fails, the two engines can continue to operate. Only if an engine fails THEN then driveshaft becomes a single point of failure which can be critical. Or do I oversee something here?

@U_flow it depends on the failure mode - if the shaft snaps cleanly and the two parts continue to rotate freely then all is well but if the fragments flail around causing collateral damage then it could be much worse.  And if part of the mechanism were to seize then it’s likely to end badly for both engines/rotors.

Yes, the V-280 Valor is equipped with a driveshaft which distributes the engine power from the nacelle with the inoperative engine to the one which still does. This system allows the aircraft to continue to operate in case of a single engine failure (but obviously not both :D).
A similar approach was taken with the predecessor of the Bell V-280 Valor, the Bell Boeing V-22 Osprey. A nice aviation stackexchange question about this system can be found here.
In this source is is stated, that this system also contributed to the design decision of the V-280 to have fixed nacelles and tilting rotors:

The straight wing also greatly simplified the mechanisms needed to connect the rotors with a driveshaft to keep them coordinated and, in case of an engine failure, both rotating. That, in turn, eliminated the need for a mid-wing gearbox, a device that adds 292 lb (132 kg) to the Osprey’s weight and expense to its price.

Which again emphasizes, that the driveshaft between the two nacelles distributes engine power in case of an engine failure.


text:
When you need to tell airport employees where you want to go, do they prefer it if you use the IATA code or the ICAO code?
(This is a question by a civilian, but please know that I do understand the system behind the IATA codes and the ICAO codes)
I want to know, when I go to an airport to fly somewhere and want to use airport codes instead of airport names, if the airport employees prefer you to tell them the IATA code or the ICAO code to inform them of the location where you want to go

Not an airport employee so just a comment, but I would suggest IATA every time. It's what the airlines themselves use, after all. (Hence the name.)

What 4-letter code do you use if you want to tell an airport employee where you'd like them to go?  ;)

ENVA is located right next door to Hell after all...

IATA codes. That's what the airlines use, and what will be on your ticket and baggage tags.
The name of the airport is also acceptable, of course.


text:
Distinguish between planes and helicopters with ADS-B messages
Is there a way to reliably distinguish between planes and helicopters using an ADS-B receiver only (no aircraft database)?
I think of using heuristics like "if ground speed > 250 mph then it's a plane", but I would like to find some good heuristics to be sure to identify helicopters correctly.

ADSB transmits wake vortex information, and TC=4,CA=7 is for rotorcraft.
This is called the "emitter category" by ICAO and some other sources, a rotorcraft is A7.


text:
Detecting plane or helicopter landing using ADS-B messages only
Let's suppose I only have an ADS-B receiver.
I want to detect aircrafts landing at the receiver's location, and estimate the time they will take to land.
What would be the most reliable way to perform that estimation?
I can use the ground speed and baro rate if they are available, but I've seen that the baro/geom rate is often not available in the ADS-B messages.
Estimating the horizontal/vertical velocities from positions is not reliable (see Why does ADS-B broadcast velocity?).
What piece of information can I reasonably assemble from ADS-B data only?

What is the time accuracy you need to achieve? Are the aircraft landing at an airport?

@DeltaLima I should try to detect aircrafts trying to land at a location 5-10 minutes before they land. No real need for time accuracy. Not necessarily at an airport.

So you have a location, that isn't necessarily a runway, and you want to know 5-10 minutes beforehand that an aircraft is going to land there, yes? Since "mind reading" is beyond the scope of ADS-B (and any other known technology), what data to you believe could tell you that far in advance that an aircraft will be landing there, as opposed to just flying past?  With good position/velocity data, you could note that an aircraft is on course to fly near your position, but until they commit to landing there - well within 5 minutes of actual landing, in many cases - "near" is all you can know.

"Estimating the horizontal/vertical velocities from positions is not reliable" actually in your case the mean velocity from position is more useful than the Doppler instantaneous velocity from the GPS. Build a simple Kalman filter to predict the landing time or use a one from a library. More.

@mins interesting suggestion to use the Kalman filter. My purpose is for turning on landing lights, so, for answering Ralph J considerations, it's better to turn on lights one once more rather than once less.


text:
How do hybrid control surfaces work?
I am trying to build a self-stabilized model rocket with actuated fins. Suppose the rocket has 4 fins. The diagonal fins must rotate together to control the yaw/pitch motion, and all the fins are tilted to control the roll motion. However, when both the yaw/pitch motion and the roll motion of the rocket need to be stabilized, how should the tilt angle of each fin be determined? I believe this is similar to a V tail, where both the yaw motion and the pitch motion depend on the same pair of control surfaces.

You will need to combine the two inputs somehow.
The simplest method is to simply add the two inputs for each fin together. For instance, let's say your control software commands the yaw fins to deflect 2 degrees to the left, and the roll fins (i.e. all of them) to deflect 3 degrees clockwise. Then, you'll wind up with one yaw fin deflected 5 degrees, and one deflected -1 degrees. This is usually good enough for a simple model rocket.
A more comprehensive solution would require either calculating, or empirically testing, how much force each fin creates at each angle of deflection. Then, you add the desired force for each fin together, and then calculate the desired angle of deflection based on that. This is usually more stable with less chance of oscillation, but does require more work to set up.


text:
Can MH370 InmarSAT data fit an early water landing?
When I see the predicted path of MH370 based on InmarSAT-3F1 data, it is curved and follows nearly equidistant from the satellite itself. Is possible that the plane did a water landing at 18:45 UTC and slowly sank.  Can MH370 InmarSAT data fit to an early water landing?  After all, the data simply suggests that the distance between the aircraft and the satellite is getting larger. If the plane's antennas are submerged, how does this effect the transmission?
Notes/Assumptions:

The best estimated path assumes the plane is flying at a cruising speed of 400-450Knts after 18:45 UTC.  The varying BFO values indicate the aircraft was moving at speed.   I'm suggesting it lands in water and floats for several hours to possibly days.
The uplink for InmarSAT-3F1 is C-Band which is 4GHz to 8GHz.  Needs to be pointed to the satellite and requires the aircraft's navigation system to be operational.
Victims mobile phones ringing - Many relatives were able to hear a ringing tone for up to four days after the crash.
The pings from the plane to the satellite stop after about 8hrs which fits well to the amount of potential flight time the plane could stay in the air for. This one is tough to ignore.  It fits nicely.  But we will assume the plane has enough battery backup to send a signal for an additional 6hrs after the water landing. This of course would be a super big coincidence.  But the odds are equivalent to a one outer on the river.  Certainly not impossible.

So we are assuming a soft water landing near where the radar return of MH370 "suddenly" stopped.  This is at coordinates 6°36'28.75"N and 96°18'38.22"E.  This is about 70 miles away from the nearest shoreline. The plane's InmarSAT SDU transponder could still remain locked to the satellite and send out signals.  Let's assume that the plane is mostly intact and taking on water slowly. Backup power cuts out after 6hrs.  Over the course of several days, the plane drifts towards land with just enough reception to initiate connection.  Reference 13 suggests up to 40 miles over open water maybe possible.  At somepoing soon after, the plane sinks.  It is still almost entirely intact minus a few parts which float off and eventually are found on distant shores.

References:

https://personal.utdallas.edu/~zweck/MH370_UNM.pdf
https://arxiv.org/pdf/1702.02432.pdf
https://www.hydro-international.com/content/article/underwater-electromagnetic-propagation
https://www.asiasat.com/news/blog/wet-antenna-attenuation-techniques-mitigate-ultra-high-rain-fade-c-band-uplink
How does one interpret the satellite data on flight MH370?
Why do MH370 Indian Ocean INMARSAT distances increase when they should decrease?
https://www.thesun.co.uk/news/8741602/mh730-mystery-satellite-turned-on-vanished/
interactive.satellitetoday.com/inmarsat-exec-talks-about-operators-role-in-search-for-mh370/
mh370.org/mh370-fuel-endurance/
mh370.org/mh370-penang-diversion/
mh370.radiantphysics.com/2018/04/11/the-civilian-radar-data-for-mh370/
en.wikipedia.org/wiki/Malaysia_Airlines_Flight_370_satellite_communications
mobiletechaddicts.com/cell-phone-signal-underwater/

This seems more like a question about radio transmission than aviation. You might get a better answer at https://ham.stackexchange.com/

Some of the things that need to be considered here is the index of refraction of air, humid air and sea water at 4GHz.  Basically, the transmission will be slower as the EM wave travels through higher index of refraction.   Not only that, you must take into account the full amount of each material.  For instance, at 40,000ft, there is far less total atmosphere for the wave to travel through. Thus the duration of travel will be less.  Which will make the craft seem closer to the satellite than it really is if you don't take this into account. Submerged meters in seawater is a big difference.

@Mark P. The frequency offset used in the analysis is a measure of relative speed of the satellite and the aircraft. The refraction index does not have a significant impact on the measured frequency offset. The refraction index relates to the speed of travel of signal, not its frequency. Furthermore, submersion in sea water attenuates the signal so strongly, that I doubt there is any communication possible with the geostationary satellite if there is more than a few centimeters of sea water involved.

@DeltaLima BFO is a complex calculation.  Even the raw data that In InmarSAT passed along has stuff baked into it.  I'm still studying it.   Some additional interesting references: https://arxiv.org/pdf/1702.02432.pdf  https://www.atsb.gov.au/publications/2014/mh370-burst-timing-offset  http://jeffwise.net/2016/09/25/assessing-the-reliability-of-the-mh370-burst-frequency-offset-data/


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Headset for both plane and heli: buy for plane, use adapter for heli, or vice versa?
I fly both planes and helicopters and am about to buy a Bose A20. I'm wondering if I should get the helicopter version with a U-174 plug and a coiled cable, or the plane version with a dual plug and a straight cable. (In both cases of course I'd also get an adapter to go from one plug to the other.) Does someone have recommendations as to which way to go? E.g. is the coiled cable a hassle in the plane, or a feature? Is one of the two plug types better, or more impaired by the adapter?
Thanks!

There's no issues with the adapter either way, as long as it's good quality. The helicopter versions are coiled for a good reason: to keep it out of your way, so I would buy the helicopter version. In most fixed wing airplanes the coil will be absolutely fine, if there's not enough reach you can get extenders pretty easily.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Is responding to an internet forum post looking for a scenic flight "holding out"?
Suppose I'm a private pilot and see a post in a closed group on Facebook where someone is looking for a scenic flight around my area. If I offer them a flight with me, sharing costs equally, am I holding out? Does it matter if I know the person vs. if I don't? Thanks!

Probably. Read the Haberkorn opinion from the FAA. It actually contains a Facebook example.

This is almost certainly holding out: someone asks for a flight and you agree to provide it. You weren't going to make the flight until the person asked, and you're receiving compensation for it. This is from the Bobertz interpretation:

Absent a bona fide common purpose for their travel, reimbursement for the pro rata share of operating expenses constitutes compensation and the flights would be considered a commercial operation for which a part 119 certificate is required.

The Haberkorn interpretation that acpilot mentioned is somewhat similar to your scenario in that it talks about using Facebook but with the difference that the pilot advertises for passengers, rather than passengers asking for pilots. You can read the whole thing and although the FAA said they can only comment on specific cases, not general examples, they did make clear that they're concerned about situations where:

there is an interest in carrying passengers with whom there is no previous personal relationship and that the offer to accept passengers is being made to the general public

But even if you already know the individual, it's probably still a commercial operation.

So in summary, two people knowing each other and sharing costs equally constitutes a commercial operation?  Only in aviation...  FYI, neither link works.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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How can I tag along on an Atlantic crossing?
I'm a private pilot based in the US and interested in tagging along on a North Atlantic crossing. I have too little experience to dare to attempt it myself, but I'd love to experience it. Does anyone know if there is a way of making that happen? E.g. are there pilots who take guests along? Or are there legal or insurance impediments?
Many thanks for any information!

Just go to the airport and hold out your thumb and hope things work out.

Contact companies that do ferry/delivery services.  You'd probably have to pay for your own immersion suit, IF someone was willing to let you come along.  Yes insurance will also probably be an issue.


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Why does a sharp leading edge result in a larger pressure peak?

I know in the real world the sharp leading edge (LE) will results in a larger pressure peak compared to a large radius LE.

And that there are three types of stall.
But I want the explanation in theory. I can't overcome the problem why the smaller the radius of the leading edge, the larger the pressure peak will be.

Do you have some reason to think it should be otherwise?

Envisioning the air flow at 15 degrees AoA is very helpful in understanding the negative pressure distribution on the upper wing.  Smoke in wind tunnels is used so the airflow pattern can be seen.  Yes, it is illogical for a thinner wing to have a higher pressure peak,
but notice the TOTAL negative pressure area of both 4408 and 4415 is very close.
What is going on here is the larger frontal area of the 4415 creates a larger, but not as low pressure bubble at 15 degrees AOA (look at graph) than the 4408.  Try superimposing the two airfoils at 15 degrees angle of attack and also throw in a flat plate too.
Remember, a higher pressure peak does not mean higher total negative pressure.  It means, like @Jan Hudec says, air molecules, which have mass and velocity (inertia), can not turn a sharp corner as easily as a small turn.
Imagine 20 race cars going down the track together.  They will make a small turn together, but if too fast into a very sharp curve, there will be a vacuum peak (lack) of cars on the near side of the turn!

Thank you for your wonderful answer.

To avoid the leading edge, the air is initially deflected upwards. Due to inertia, it would continue upwards. However, viscosity prevents it from having sharp shears of velocity, so the air just above the wing gets pulled out, creating the area of reduced pressure that pulls the oncoming air to turn around the wing—the suction peak. Sharper leading edge means the angle between the air and the curving surface grows faster, so it needs deeper suction peak to keep the air attached.
Since the viscous forces are only so strong, they can only make the air follow a so sharp curvature. As the angle of attack increases, the air flowing over the upper surface has to navigate around more of the leading edge. With sharper leading edge the angle it has to turn increases faster, so it will start to separate earlier.
The trailing edge stall is also related to viscosity. It prevents velocity discontinuity also at the wing surface, so near the surface there is boundary layer where the stream velocity changes from zero to the free stream velocity. This boundary layer grows in thickness, more at higher angle of attack, and when it becomes so thick that the viscous forces are insufficient to keep it moving, a separation bubble creeps in from the trailing edge. This is why wings with longer chord stall at lower angle of attack. Even if it is the same profile scaled up.
This also explains how vortex generators increase the critical angle of attack: turbulence mixes the boundary layer, increasing the velocity in it even if the average aft component does not increase, which prevents it from stagnating and separating completely.

NACA nomenclature 4408 means 4% camber at 40% chord 08% thickness.  Would a 4415 have the same top profile (relative to chordline) if they were superimposed?

Thank you for your wonderful answer.


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Would a coaxial propeller with the second prop slower and larger improve speed?
Looking at various propeller designs, I started wondering about these limitations:

Large propellers at high speeds have tips going supersonic
Scimitar props attempt to keep prop tip from going supersonic
Counter-rotating props absorb more power at smaller sizes

This made me wonder if a coaxial prop had a larger prop spun at slower speeds and a smaller prop spun at high speeds would increase thrust better, or do other designs already deal with that problem?

This answer was proposed for a different question, but it seems to have some useful crossover: https://aviation.stackexchange.com/questions/53422/what-s-the-thrust-of-two-stacked-propellers/53440#53440

The larger propeller would dictate the Mach limits of the whole configuration, so unequal propeller sizes would be slower than otherways identical ones with equally sized propellers.

I think what you'd ideally want to do is to have a fast-rotating "inner" propeller and a slower "outer" one, or actually a propeller whose RPM gradually decrease from the inside to the outside. Although that would look cool, it doesn't work with non-fluid materials for the rotor :o) So, the next best thing is to have two behind each other, where the downstream one spins a bit slower.
How I understand your proposition
Counter-rotating props do have the advantage that the second one can "straighten" out the flow from the first, effectively working similar to (but not entirely like) a rotor/stator couple in a turbojet. This means they can transfer more power per rotor area than single rotors. By putting the slower-spinning one behind, this couple will leave less swirl in the flow than the other way round, which sounds like a good idea to me.
Difficulties with regular counter-rotating rotors
In a counter-rotating rotor couple, the second rotor is exposed not just to the mean swirl coming from its upstream partner (for which it is designed, and which can increase its effectiveness), but also the pressure signatures and wakes coming off each blade. Every time a blade passes through one of these, the stagnation pressure at its leading edge goes down very sharply and then increases again. That makes a lot of noise, and also causes vibration in the blade, which then has to be made strong enough to deal with this. This is also the reason why most such rotors have swept blades like these:
Antonov AN-70 with swept counter-rotating props; image from https://wordlesstech.com/revolutionary-airplane-propeller-action/
This means that at no moment will the entire blade of the downstream rotor be in the wake of an upstream blade, but it will gradually pass through it. Still the blades have to be quite sturdy, which also prevents them from being very long. That, in turn, limits the ability to increase efficiency by making long, slender blades which produce less thrust per rotor area but make up for it in radius (like wind turbines, for example). That's one reason why most such configurations these days are seen in military airplanes where large elegant propellers would reduce maneuverability but lots of thrust is needed, pronto. Also, these machines are built very robustly anyway, so the added vibrations can be dealt with. In a passenger aircraft, the noise alone would be hard to sell to airlines, but developing the mechanical components would also not be easy. At the moment it seems that only Antonov has experience with large configurations of this type.
That said: These concepts keep cropping up, and it seems as if it should be possible and efficient to use counter-rotating rotors with strongly swept blades to replace turbofan engines at speeds that would seem a bit high for regular propellers (but maybe a bit low for turbofans).
Here are two sources I just found which give you some idea of what the flow looks like, and what types of aerodynamic and acoustic problems the designers have to content with:

A 2014 paper summarizing design and research activities
Slides from a 2010 presentation on simulating the flow, and on efficiency and acoustics -- very nice pictures, and a good way to get an idea of what propeller flow looks like.

The problem with having a larger downstream prop
You can also see in the references above that usually the downstream blades are actually shorter than the upstream blades. The reason is that otherwise the tip vortices from the upstream propeller would hit the downstream one, and that's acoustically, aerodynamically and structurally not nice. Even without those vortices, it would make sense because the downstream prop gets more vibrations either way, so it has to be sturdier (shorter, thicker blades will help!).
I think some of the existing configurations (the AN-70 above for example) do have equal blade lengths, but they are very much known for being loud, and any new civil application cannot have that.
The most recent example of a counter-rotating open fan that has actually been built and tested is the SAFRAN open rotor:
photo by Eric Drouin / SAFRAN
SAFRAN claims that the design can fly at Mach 0.8 and is about as loud as a comparable turbofan engine from the previous generation (i.e. still louder than the latest turbofans, but not horrible, either). As you can see, the downstream rotor is also shorter in this configuration.
Noise, by the way, is also one reason why most configurations under discussion (with some exceptions, because there always are...) are rear-fuselage mounted rearward-facing rotors: That puts the main source of noise a good way downstream of the passenger cabin, instead of right next to their windows (which is loud enough with regular props already, as any passenger in a Saab 2000 can confirm). It also has the benefit of preventing accidents where a blade breaks off and hits the passenger cabin ...

Aerosila reduced the interference noise by using different numbers of blades on the SV-27, eight on the front propeller and six on the aft propeller.

yes, not using equal number of blades is the first thing to do. Ideally, all axial compressors/turbines should also be using different prime numbers for each stator/rotor row, to prevent two pairs of blades from being in the same relative position at the same time. Of course, in a multistage compressor, you run out of prime numbers quickly, but for counter-rotating open rotors, you'd need a really good reason to use equal numbers. I actually wonder why Aerosilia didn't go with 8 vs. 7  or 7 vs. 6 instead, would seem smarter.

Assuming that your question does not concern situations like the AN70 where there is so much power for the prop to absorb that both props need to be as large and solid as possible, what we are looking for is the most efficient thrust distribution through the prop disc. You want a smooth distribution as you would for a wing.
The thrust developed on a rotating prop blade is lower at smaller radii where relative speed is slower. Most of the thrust of a typical prop blade is in the outer third. This looks like there might be an opportunity for another smaller blade to provide additional thrust toward the center in order to get a more uniform distribution across the prop disc.

However, you don't really want an elliptical lift distribution on a blade as you might see on a wing for the same reason that the inner prop doesn't produce as much thrust; the inner prop doesn't travel as far and has less volume to fill behind the prop.
What you really need to look at is the slipstream. Here you see that the 'hole' in the slipstream is much smaller than the blade thrust distribution would indicate, and much of that is filled by the engine nacelle. There is perhaps a five percent improvement possible. A pusher prop might be different but even there the center quarter of the prop is only six percent of its area, limiting its potential for improvement.

The conclusion is that the theoretical efficiency improvement does not justify the weight and complexity of a coaxial prop where excess power is not a constraint.

Coaxial, contrarotating props are one way to improve engine efficiency over a single propeller.  When doing static or bench testing with no airflow into the props, the front prop will carry more load than the back one (which may have lead to this question).
Fixed pitch propellers "unload" a bit once the plane gets moving.  A distinct increase in rpm can be heard with models.  This effect is also called "windmilling".
With a contra rotating prop, the loads on the front and back props will therefore even out more in flight.
An increase in weight and complexity of the gearbox has caused many to shy away from this design, but it is an interesting study as powerful electric motors with much higher rpm ranges are coming into use.

"the front prop will carry more load than the back one " - How about some supporting reference?

Increased load of leading prop on static bench test may be due to increased drag/ AoA effects.  A coaxial prop could be load tested in a wind tunnel to give  power draw curves of both props at flying speeds.  Individually powered electrics mounted in a coaxial configuration (of varying distances apart on varying airframes (such as Dornier DO 335 "Pfeil")) would make a great study,

Windmilling is what happens when the engine quits and airflow turns the prop.  (Like a…windmill!). It doesn’t happen if the engine/prop is producing power.

@MichaelHall perhaps a better term would be more appropriate.  The point is that bench testing a co-ax (or any prop) at a given RPM can produce different thrust results then in the air due to changes in relative wind.

Contrarotating propellers may have a better efficiency than a single propeller, since the propwash is 'straightened out', the tangential component of the airflow being much reduced. Besides, and for very high engine powers, those props have also the advantage of being able to absorb a much higher power. But that's all...

How about some supporting reference?


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Can you fly to a U.S. border airport without any paperwork?
In most cases, one can fly VFR in the U.S. without any flight plan. However, if you are crossing an ADIZ, it is my understanding that you have to file a DVFR flight plan.
Consider an airport such as KDGL (Douglas Municipal). In theory, it looks like one could fly to the airport from the north without ever crossing into Mexico, but as soon as you need to do a go-around you would be crossing the U.S. ADIZ into Mexican airspace (and crossing back into the U.S. when returning to the airport).
Do you need to file a flight plan (or any other paperwork or paying overflight fees) when flying to such an airport? Do you also need to file a flight plan if you are simply flying the pattern at the airport?

In the case of KDGL you need to keep your flying tight and plan your go around to avoid busting the AIDZ. Interestingly there is nothing in the AF/D about this and not any great written info on it. But there is a discussion you can look over here where someone reached out to the FBO about it and the answer was:

So I decided to take one for the team and bite the bullet and call the
FBO to find out... he said that it's usually not an issue... most
pilots do keep their pattern tight, and they cross over if they need
to all the time. The FBO does get calls from Riverside, CA customs all
the time to check traffic, but there's usually no issue with it.
BTW... it is right traffic for 3 and left traffic for 21.

You may also want to checkout this related IFR scenario where effectively the appropriate ATC facilities coordinate as needed.


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Is there a crosswind that would perfectly cancel out the "torque" of a C172 on takeoff?
We all know that on takeoff you need right rudder in order to balance out the "torque" of the propwash hitting the tail.  Is there an exact crosswind that would cancel out this torque?

NOTE: This is not meant to be a scientific analysis. In older 172's such as a 172M, when there is a 2-3 degree crosswind perpendicular to the path of the airplane it seems to cancel the torque out on departure. In the newer models like the 172S, it is more like 4-5 knots.

Fascinating experiential answer!

It is reasonable to believe, (as it is with most hypothetical questions of this sort...) that the answer is yes.  However, there are a few things to consider:

It would need to vary to the degree that the throttle is advanced.

It would need to vary during the acceleration as aerodynamic effects change.

So, if you think you might be able to listen to ATIS and discern that you won't need rudder on the takeoff roll, it isn't that simple.

Then there was gyro precession effects on taildraggers.  When I was  part owner of a C-180, when it was on wheels, if you tried to raise the tail early, with a bit too much gusto, the propeller precession would swing it left pretty hard.  If there was a strong left xwind, you'd find yourself running out or rudder.  Much safer to hold the tail down with elev and let it come up gently after past about 30kt.

Good point.  I consider adding this to my answer, but the OP asked specifically about a 172.

Yeah I know I was just story telling.

Yes, but not really.
If the the 172 were traveling at a constant velocity and with a constant prop speed (rpm), then there would be a linear combination of crosswind direction and wind speed that would exactly cancel out the torque.  For example, a 90 degree crosswind would need a lower velocity to counter the torque, whereas a narrow angle crosswind would need a much higher velocity.
But on takeoff, the aircraft velocity will be increasing over time.  Whatever crosswind direction and velocity numbers you chose which fit the example above would not work for all other aircraft velocities.  As the velocity increases, the effective force on the vertical stabilizer (and fuselage) from air will change.
To make this easier to imagine, consider a craft moving at 500kts with a 20kts 90 degree crosswind vs a craft moving at 5kts with the same crosswind.  The faster the craft is moving forward, the less off-axis wind forces will matter.
So when the aircraft is motionless, and the engine is set for takeoff power, the crosswind direction and velocity combination that will perfectly counter the torque will be one set of numbers, but as the aircraft increases speed the perfect counter numbers will change.  Ignoring the complexities of fluid dynamics, this could be represented in a 3D graph (crosswind direction, crosswind velocity, and aircraft velocity).  It would look something like a tilted triangle.


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Relationship of propeller RPM, blade pitch, and drag at very low blade pitch angles
I'm researching the 2016 crash of a PIA ATR-42 in Pakistan, the report on which was recently released. Link to the report: https://www.caapakistan.com.pk/Upload/SIBReports/AAIB-386.pdf
During the incident, a series of malfunctions caused the propeller on a failed P&W 127M engine to not feather although it was commanded to do so. The overspeed governor was also not working; as a result, blade pitch began to decrease and RPM increased past the normal overspeed stop (>120%) causing massive drag.
However, at a certain point, the RPM suddenly dropped (<25%), causing a massive decrease in drag. This was critical to the sequence of events as the pilots were standing on the rudder to compensate for the drag, the sudden disappearance of which caused them to overcontrol and lose control of the aircraft within a couple of seconds.
What I don't get is why the trend toward increasing propeller RPM as blade pitch decreases would suddenly reverse if the pitch gets low enough. This is what the report says:

"If  blade  angle  decreases below  a  certain  angle (depending  on  the  Np  and  the  power  absorbed  by  the  engine),  the propeller  in  wind-milling  does  no  longer  generate  power  but  needs power  to  stay  at  the  same  rotational  speed.  With  a  failed  engine,  the decrease  of the blade pitch  angle below  a  certain  threshold  leads  the propeller speed to decrease."

"There was sudden drop in Np-1. It  was  because  of  the  reason  that  the  blade pitch  angle (most  likely) decreased  further  up  to  a point  where  the  power  generated  by  propeller  was lower than power absorbed by the engine and it may have later  moved  to  a  stable  physical  position.  Np-1 decreased below 25% and then stabilized lower than 5%, with blade pitch angle close to low pitch in flight. This  sudden  decrease  in  the  Np-1  resulted  in sudden depletion of large amount of drag."

What do they mean by "the propeller  in  wind-milling  does  no  longer  generate  power  but  needs power  to  stay  at  the  same  rotational  speed"? If it needs power to stay at the same rotational speed, where was that power coming from when it was spinning at 120% np for several minutes with the engine shut off? And if RPM remains high, why would a further decrease in blade pitch cause less "power" to be "absorbed" by the engine? What exactly do they mean by "power" in the first place?
Furthermore, if this drop in blade pitch towards "low pitch in flight" resulted in a decrease in RPM and a commensurate decrease in drag, the next part of the sequence of events also needs clarification. After recovering from the snap roll and dive, the crew were faced with a situation where the propeller on the failed engine was rotating very slowly (under 5% np) with blade pitch at or slightly below zero degrees, but was also causing even more drag than when it was at 120%. How is this possible, if just moments earlier, blade pitch decreasing toward that value caused drag to decrease?
Any help would be appreciated. Thanks.

it should be like this:
As the pitch of a windmilling propeller decreases (and its speed increases), it's absorbing an increasing amount of power from the air (hence, this increase in absorbed power is directly correlated to an increase in propeller drag).
At a certain point, the pitch becomes so low that the blades start to stall and the propeller RPM start to decrease. This, counter-intuitively, produces a decrease in propeller drag.
Once the propeller is at max fine pitch, with the blades mostly stalled and a very low (or zero) RPM, it is neverthless producing a lot more drag than in the feathered position, although less drag than it was at max RPM (>120% in the above case).
So, probably the pilots were still able to momentarily control the aircraft even with the propeller at >120% RPM because, in the initial stages, they still had enough airspeed and hence enough directional control.
In the later stages, with the propeller at min pitch and low RPM, the airspeed was progressively depleted so that they lost directional control, even though the propeller drag was at that point somewhat lower than it was at >120% RPM.
To confirm this, I tried the scenario in X-Plane 12, which should have a decent simulation of propeller dynamics. I used a B90 instead of an ATR42, but the results seem to confirm my explanation. You can see in the graph below propeller RPM (grey line) and propeller drag (magenta line). After I failed the propeller governor to fine pitch, the RPM start rapidly increasing along with the drag, both of them reaching a peak. Then, the pitch becomes so low that the RPM starts to decrease, and the drag settles at about 60-70% of peak drag.